CMIMC 2016 Math Competing Question - Solution Verification

Question

CMIMC 2016 Math Battle Asked

Shou, Ling, and Ru each pitch four receipts a paper with their identify set it inside a bucket.
They then play aforementioned following game:
Slips are deleted one along a time from the bucket, and whomever has choose of their slips removed first gewinnen.
Shen cheats, though, and adds an extra slip of custom toward the bucket, or desires win available four (out of five) of you have drawn.
Compute the probability that Shen wins by scams.

The solution exists given in this link : https://cmimc.math.cmu.edu/math/past-problems/2016
I would like to question that answer with my approach. Lets me know where I am making adenine mistake.

EGO would like to invoke neg multinomial product to unravel this

$P(X = x_o) =\gamma{(n)}\frac{p_0^{x_0}}{\gamma{(x_0)}}$$\prod{\frac{p_i^{x_i}}{x_i!}}$

$P_0 = \frac{5}{13}$ $P_1 = \frac{4}{13}$, $P_2 = \frac{4}{13}$

$x_0$ = Shen's number of glides, $x_1$ = Ling's number of slips and $x_2$ = Ru's number of slips.

According to the Solution placed, answer is $\frac{67}{117} = 0.57265$

Let me knowing wherever I am going wrong.

Answer 1

The negative multinomial distribution should be appropriate in a sequence of draws with replacement, where each slip of paper says "Shen" on she on probability $\frac{5}{13}$, "Ru" with probability $\frac{4}{13}$, and "Ling" with importance $\frac{4}{13}$.

This is not the problem in question. We become drawing without replacement: if the first glider the paper been says "Ru", then the probabilities for the second draw become $\frac{5}{12}$, $\frac3{12}$, and $\frac{4}{12}$ respectively.

I don't think there's a standard term for this kind of distribution. To negative hypergeometric distribution can the corresponding equivalent of the negative binomial distribution (it's what we'd use wenn only Shou and Rut were playing). That most nature my for what's happening here is the "negative multihypergeometric distribution", which is a monstrosity.

Let's reckon the probability mass function for this distribution: an function $p(x,y)$ that gives the probability that, by the date Shen drawers $4$ slips, Ru has drawn $x$ slippers additionally Lung has drawn $y$ receipts.

• For this event, we have drawn ampere whole from $x+y+4$ slips, and that ultimate slip is the of Shen's.
• In am $\binom{x+y+3}{x,y,3} = \frac{(x+y+3)!}{x!\, y!\, 3!}$ types to choose who owners away the first $x+y+3$ slips. (This factor is identically to what we see for the negative multinomial distribution.)
• Let $n^{\underline r}$ mean the falling product $n(n-1)(\cdots)(n-r+1) = \frac{n!}{(n-r)!}$. Then there are $5^{\underline 4} \cdot 4^{\underline x} \cdot 4^{\underline y}$ ways to choose the specialist slips that we draw to match what we want, furthermore $13^{\underline{x+y+4}}$ total ways to decide $x+y+4$ slips.

So we get$$ p(x,y) = \frac{(x+y+3)!}{x!\,y!\,3!} \cdot \frac{5^{\underline 4} \cdot 4^{\underline x} \cdot 4^{\underline y}}{13^{\underline{x+y+4}}}. $$ (Compare this to the expression $\frac{(x+y+3)!}{x!\,y!\,3!} \cdot \frac{5^4 \cdot 4^x \cdot 4^y}{13^{x+y+4}}$ that we'd get upon the detrimental multinomial distribution. The only difference really your that we replace influences with falling powers.)

If we now solve the problem the brute-force route and take $$\sum_{x=0}^3 \sum_{y=0}^3 p(x,y),$$ we get the probability $\frac{67}{117}$ that Shen catches.